2020 AMC 10B Problem 18

The urn ends with three red and three blue balls exactly when the four draws contain two red draws and two blue draws. There are $\binom42=6$ possible color orders of this type. For any fixed order with two red draws and two blue draws, the probability is $$\frac{1\cdot 2\cdot 1\cdot 2}{2\cdot 3\cdot 4\cdot 5}=\frac{4}{120}=\frac{1}{30},$$ because the first and second draws of each color have numerators $1$ and $2,$ while the total number of balls before the four draws is $2,3,4,5.$ Thus the desired probability is $6\cdot\frac{1}{30}=\frac15.$ Thus, the correct answer is B .